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Alex, Bret, and Chloe repeatedly take turns tossing a fair die. Alex begins; Bret always follows Alex; Chloe always follows Bret; Alex always follows Chloe, and so on. Find the probability that Chloe will be the first one to toss a six.

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marked as duplicate by Matthew Conroy, copper.hat, Stefan Hansen, Sami Ben Romdhane, Michael Hoppe Feb 21 at 7:41

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Any your ideas? –  sas Feb 21 at 5:58
    
What is the probability that Chloe rolls the six on her first turn (with neither Alex or Bret rolling a six before her)? What's the probability that she rolls the first six on her second turn? What's the probability that she rolls the first six on her $n$th turn? Are these events disjoint? –  Sammy Black Feb 21 at 6:08
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3 Answers

A natural approach involves summation of series. We look at the problem another way.

Let $p$ be the required probability. The event "C is first to get a $6$" can happen in two ways:

(i) In the first $3$ tosses, the results are not $6$, not $6$, $6$. The probability of this is $\left(\frac{5}{6}\right)^2\cdot\frac{1}{6}$.

(ii) In the first $3$ tosses, the results are not $6$, not $6$, not $6$, but C ultimately is first to get a $6$. The probability of this is $\left(\frac{5}{6}\right)^3p$.

It follows that $$p=\left(\frac{5}{6}\right)^2\cdot\frac{1}{6}+\left(\frac{5}{6}\right)^3p.$$ This is a linear equation in $p$. solve.

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The third player in line wins if the first player fails to roll a six ($5/6$) and then the second player in line wins. So $p_3 = \frac{5}{6}p_2$. Similarly, $p_2 = \frac{5}{6}p_1$. Finally, the first player in line wins if he rolls a six immediately ($1/6$) or if he fails to roll a six ($5/6$) and then the third player in line wins: $$ p_3 = \frac{5}{6}p_2 = \frac{5}{6}\left(\frac{5}{6}p_1\right)=\frac{5}{6}\left(\frac{5}{6}\left(\frac{1}{6}+\frac{5}{6}p_3\right)\right)=\frac{25}{216}+\frac{125}{216}p_3. $$ Solving this gives $$ p_3=\frac{25}{216-125}=\frac{25}{91}; $$ then $$ p_1=\frac{1}{6}+\frac{5}{6}\cdot\frac{25}{91}=\frac{91+125}{6\cdot 91}=\frac{36}{91} $$ and $$ p_2=\frac{5}{6}\cdot\frac{36}{91}=\frac{30}{91}. $$ These satisfy $p_1+p_2+p_3=1$, as they must.

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Let $S$ be the event that a six is thrown, and $\bar{S}$ the event that a six is not thrown.

Then the patterns with which Chloe wins are $(\bar{S} \bar{S} \bar{S}) ^k\bar{S}\bar{S} S $, hence the probability that she wins is given by $\sum_{k=0}^\infty ({ 5 \over 6})^{3k+2} {1 \over 6} = { 5^2\over 6^3} { 1\over 1-({5 \over 6})^3}$.

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@AndréNicolas: Oops... –  copper.hat Feb 21 at 6:09
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